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$4^2+4^2+4^2+4^2=2^n$ what is the value of $n$

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Find derivatives using the product rule I

Full solution

Q.

4^2+4^2+4^2+4^2=2^n

what is the value of

n

Calculate $4^2$ : First, let's calculate the value of $4^2$ . Since $4^2$ means $4$ multiplied by itself, we have $4 \times 4$ , which equals $16$ .
Add $16$ s: Now, we have the equation with four identical terms added together: $16 + 16 + 16 + 16$ . Adding these together gives us $16 \times 4$ , which equals $64$ .
Express $64$ as $2^n$ : The equation now is $64 = 2^n$ . To find the value of $n$ , we need to express $64$ as a power of $2$ . Since $64$ is $2$ raised to the $6$ th power ( $2^6 = 64$ ), we can say that $n$ equals $6$ .
Determine $n$ value: We have determined that $2^n$ is equal to $64$ when $n$ is $6$ , so the value of $n$ in the equation $4^2+4^2+4^2+4^2=2^n$ is $6$ .

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